From 6.2:
4. (a) The first digit can be anything but 0, so we have 9 choices. The second
digit can be anything but the first digit, so 9 choices. The last digit can be
anything but the first two, 8 choices. The total is 9 × 9 × 8 = 648.
(b) There are five choices for the last digit (1, 3, 5, 7, 9). The first digit can
be anything but 0 or the last digit so there are 8 choiced. The second digit
can be anything but the other two, so 8 choices. The total is 5 × 8 × 8 = 320.
10. We have 8 choices for each of the first two digits, 9 choices for the third digit,
and 10 choices for the last four digits. The total is 82 × 9 × 104 = 5, 760, 000.
14. (a) There are three choices for the first leg and five for the second leg, so a
total of 3 × 5 = 15.
(b) There are 15 routes to get from Cupids to Hearts Desire, as we saw in
(a). Next choose one of the 5 routes to go back from Heats Desire to Harbour
Grace, then choose one of the 3 routes from Harbour Grace to Cupids. The
total is 15 × 5 × 3 = 225.
(c) Again, there are 15 ways to get to Hearts Desire. Now we can’t repeat
the road we already used to get back to Harbour Grace, so we have 4 choices.
We can’t repeat the road we used between Harbour Grace and Cupids so we
have 2 choices. The total is 15 × 4 × 2 = 120.
From 6.3:
12. Let ai be the number of words George learns on day i, and let bi = a1 + a2 +
· · · + ai be the number of words George has learned through the ith day. Since
ai ≥ 1 for all i, we know that
1 ≤ b1 < b2 < · · · < b53 ≤ 90
and that
16 ≤ b1 + 15 < b2 + 15 < · · · < b53 + 15 ≤ 105.
These are 53 × 2 = 106 integers between 1 and 105, so by the Pigeon-Hole
Principle, two of them must be the same. We can’t have bi = bj or bi + 15 =
bj + 15 so there must be some i and some j such that bi = bj + 15 and i > j.
Then we have a1 + a2 + · · · ai = a1 + a2 + · · · + aj + 15. Cancelling like terms
gives aj+1 + aj+1 + · · · + ai = 15. The consecutive days are those between
day j + 1 and day i.
17. Divide the triangle into four smaller equilateral triangles with side lengths of
1
2 as shown below.
1
2
·T
· T
·
·T
· T
T
·T
· T
·
T ·
T
·
T·
T
By the Pigeon-Hole Principle, of any five points chosen in the triangle, two
must lie within the same smaller triangle. The furthest apart these two points
can be is the side length 12 .
23. Any number between 1 and 200 can be written as 2k a with k ≥ 0 and a odd
such that 1 ≤ a ≤ 199. The possible values of a are 100 odd numbers. If we
choose 101 integers from {1, 2, . . . , 200}, two of those integers, x and y must
have the same value of a. That is, two of our chosen numbers are x = 2i a
and y = 2j a with i > j. Thus x = 2i−j (2j a) = 2i−j y so x is divisible by y.